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16. 3Sum Closest 👍

  • Time: $O(n^2)$
  • Space: $O(|\texttt{ans}|)$
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class Solution {
 public:
  int threeSumClosest(vector<int>& nums, int target) {
    int ans = nums[0] + nums[1] + nums[2];

    ranges::sort(nums);

    for (int i = 0; i + 2 < nums.size(); ++i) {
      if (i > 0 && nums[i] == nums[i - 1])
        continue;
      // Choose nums[i] as the first number in the triplet, then search the
      // remaining numbers in [i + 1, n - 1].
      int l = i + 1;
      int r = nums.size() - 1;
      while (l < r) {
        const int sum = nums[i] + nums[l] + nums[r];
        if (sum == target)
          return sum;
        if (abs(sum - target) < abs(ans - target))
          ans = sum;
        if (sum < target)
          ++l;
        else
          --r;
      }
    }

    return ans;
  }
};

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class Solution {
  public int threeSumClosest(int[] nums, int target) {
    int ans = nums[0] + nums[1] + nums[2];

    Arrays.sort(nums);

    for (int i = 0; i + 2 < nums.length; ++i) {
      if (i > 0 && nums[i] == nums[i - 1])
        continue;
      // Choose nums[i] as the first number in the triplet, then search the
      // remaining numbers in [i + 1, n - 1].
      int l = i + 1;
      int r = nums.length - 1;
      while (l < r) {
        final int sum = nums[i] + nums[l] + nums[r];
        if (sum == target)
          return sum;
        if (Math.abs(sum - target) < Math.abs(ans - target))
          ans = sum;
        if (sum < target)
          ++l;
        else
          --r;
      }
    }

    return ans;
  }
}

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class Solution:
  def threeSumClosest(self, nums: list[int], target: int) -> int:
    ans = nums[0] + nums[1] + nums[2]

    nums.sort()

    for i in range(len(nums) - 2):
      if i > 0 and nums[i] == nums[i - 1]:
        continue
      # Choose nums[i] as the first number in the triplet, then search the
      # remaining numbers in [i + 1, n - 1].
      l = i + 1
      r = len(nums) - 1
      while l < r:
        summ = nums[i] + nums[l] + nums[r]
        if summ == target:
          return summ
        if abs(summ - target) < abs(ans - target):
          ans = summ
        if summ < target:
          l += 1
        else:
          r -= 1

    return ans